text:
How do Antishock Bodies / Cuchemann Carrots delay shock-induced flow separation?

There are several online resources that describe these aerodynamic devices as either "Antishock Bodies" or "Kuchemann Carrots" and that they are used to produce some combination of the following effects:

Delay shock-induced flow  separation
Contribute to the Area-Rule concept
Both of the above (source-dependent)

I cannot find any reading about how these devices perform the task of delaying shock induced flow separation. I am interested in how exactly these bodies interact with the combination of supersonic and subsonic flow across the chord line of a transonic wing in order to delay shock-induced flow separation to some higher Mach number than would be possible without them installed.

As I understand your question, you're familiar with the area rule. (If not, see: What is the physical-based/theoretical reason why supersonic area ruling works?)
What the area ruling does is lessen the strength of the shock wave.
And indeed yes, the stronger the shock wave, the bigger its effect on the boundary layer:

Via Delery, J.: Shock-wave boundary-layer interactions. In: Babinsky, H., Harvey, J. (eds.) Physical Introduction, Chap. 2, pp. 5–86. Cambridge University Press, New York (2011)

Explanation (emphasis mine):

6.4.3 Shock-Wave Boundary-Layer Interaction in Transonic Flow [...]
When the shock wave strength increases, the interaction between shock wave and boundary layer becomes stronger. The sonic line descends deeper into the boundary layer and a different equilibrium results. [...] The flow that goes through the two oblique shocks experiences less total pressure loss than the flow going through the normal shock wave [...] However, due to the large adverse pressure gradient under the oblique shock, the velocity at the wall quickly reduces to zero at point S. [see figure] The flow separates and a bubble of recirculating flow forms. [Source: Vos, Roelof, and Saeed Farokhi. Introduction to transonic aerodynamics. Vol. 110. Springer, 2015.]

In other words, with the > Mach 1 flow nearer the wall (wing's surface), and the boundary layer at the surface is zero velocity, you get a steeper gradient and the separation bubble.

Is this the original ymb1?

Hi @Jpe61: yes - though the answer should stand on its own. Also not here for long.

Oh. Dang, I was hoping you were back

I cannot find any reading about how these devices perform the task of delaying shock induced flow separation.

Just ask the guy who invented them (in the USA), i.e. Richard T. Whitcomb himself:
"Addition of the bodies [...] essentially eliminated the separated flow at Mach numbers up to 0.90 and greatly reduced it at higher values. This reduction in separation results primarily from the effect of the pressure fields for the bodies on the general flow about the wing, this effect being broadly defined by the changes of the localized area developments shown in figure 3. [displayed here under, Ed.] Physically, the forward parts of these bodies decelerate the supersonic flow more gradually ahead of the shock wave above the wing with a resulting decrease of the strength of the shock and the associated boundary-layer separation. [...] The bodies act effectively as aerodynamic fences. This effect is associated with the pressure gradients on the three-dimensional bodies that lead to the delay of separation on these regions as just described and not to the mere presence of vertical obstacles to the flow. Preliminary experiments of this investigation indicate that simple thin fences at the same positions as the bodies provide relatively little effective deterrent to the spanwise flow. At a Mach number of 0.95 some low-energy air moves completely over the three outboard bodies and increases the severity of separation just outboard of these bodies."

What happens is basically the same phenomenon happening on a supercritical airfoil: the particular cross-sectional shape of the previous figure (aka area-ruling) minimises the 3D pressure gradient therefore decreasing the strength of the shockwave and the chances of a detachment of the boundary layer.

Stalled swept wing at Mach 0.92: the white region is where the oil rubbed on the wing accumulated due to the stalled airflow.

The same wing at the same Mach number with Küchemann carrots aka Withcomb bodies: the stalled region is highly reduced

Whitcomb invented them? That's presumably the reason why they are called Küchemann bodies.

@PeterKämpf: shall I really specify "in the USA"? Even Küchemann himself called them "Whitcomb bodies", as visible for example in the picture of AeroAndy's answer

Well, this only shows that modesty was more a characteristic of Küchemann than of Whitcomb.

Haha, "Whitcomb who later made similar proposals"

What better source to check than Kuchemann's The Aerodynamic Design of Aircraft (1978, Pergamon Press)?

This answer would greatly benefit from better formatting. As it now stands, it is impossible for me to read.


text:
What is the best pair selection algorithm?
Distance Measuring Equipment selection criteria is said “those closest to the aircraft and which intersect at an angle between 30 degrees and 150 degrees" (also said one whose geometry is closest to 90 degrees") (Not sure range or angle is more important in operation)
So question is what should be the best pair selection algorithm for coding in operation?

When using a DME pair, the location of the receiver can be determined with some accuracy associated.
The time measurement is performed with some uncertainty leading to a slant range that is within a ring area (the double circles).

Observation
When looking at the intersection of the two rings with different positions of the DMEs:

The hatched area must be eliminated using some method
The blue area is reduced when the two DME bearings are at a right angle
As the width of a ring is proportional to the radius (the uncertainty is a percentage of the range), the smallest blue area is obtained with DME at right angle and the closest to the aircraft (last diagram).

Proposed algorithm
Divide the area around the aircraft into, say 8 sectors of 45° (the number of sectors impacts the accuracy but also the number of computations to be done, in an exponential way).

Look for the two closest stations, write down the sector(s) they belong to
Compute the intersection area
Find the next closest station D, note its sector S
If S is a sector already tested, ignore D, return to 3
Else within the already known sectors, find sector SP closest to 90° of S. Get station DC of this sector.
Compute intersection area for pair D-DC
If this area is smaller than the previous smallest area, keep it as new smallest
Repeat from step 3 until all sectors have been tested (one station per sector) or the radius has exceeded some predetermined limit.
Determine the new aircraft position using the best pair found.
Restart at step 1 using the new aircraft position (the time interval before restarting the cycle may be determined from the distance the aircraft moved. If the aircraft hasn't moved, then the next search for a pair is likely to find the same pair).

Example

Step 1: The two closest stations are in sectors S3 and S7 (fig A).
Step 2: Their intersection area is X (fig. A).
Step 3: The next closest is D3 in sector S2 (fig B).
Step 4: S2 has not yet been visited, so let's look at D3.
Step 5: In {S3, S7}, S7 is the sector closest to 90° of S2
Step 6: D3-D6 intersection area is Y (fig. B)
Step 7: Y is smaller than X, so our candidates are now D3-D6

Step 3: The next closest is D5 in sector S5 (fig C).
Step 4: S5 has not yet been visited, so let's look at D5.
Step 5: In {S2, S3, S7}, S3 and S7 are the sector closest to 90° of S5, we chose S3 because D4 is closer than D6.
Step 6: D4-D5 intersection area is Z.
Step 7: Z is smaller than Y, so our candidates are now D4-D5
Step 3: The next closest is D2 in sector S2
Step 4: S2 has already been visited, we ignore D2.
Step 3: The next closest is D1 in sector S1
Step 4: S1 has not yet been visited, so let's look at D1 (fig D).
Step 5: In {S2, S3, S5, S7}, S3 and S7 are the sector closest to 90° of S1, we chose S3 because D4 is closer than D6.
Step 6: D4-D5 intersection area is Z.
Step 7: Z is smaller than Y, so our candidates are now D4-D5
Step 3: The next closest is D1 in sector S1 (fig D).
Step 4: S1 has not yet been visited, so let's look at D1.
Step 5: In {S2, S3, S5, S7}, S3 and S7 are the sector closest to 90° of S1, we chose S3 because D4 is closer than D6.
Step 6: D1-D4 intersection area is T.
Step 7: T is larger than Z, so our candidates still D4-D5
Step 8: Our maximum radius has been reached, our pair is D4-D5

@Efes06: At step 3, by range, at step 5 by sector azimuth (as mentioned). For the latter, as sectors are, in this example, 45° wide, you use the mean azimuth of each sector, not the exact azimuth of the stations. "30/150° constraint"? If you mean D3-D6 pair in fig. B: I know I made a mistake in fig. B (I should have selected pair D3-D4 because D4 is closer than D6), but I needed to update the picture, and as it is barely visible and has no incidence on the result, I left it with the error.

source: Stack Exchange

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sort: votes


text:
How to set frequency and distance in ARINC 429?
How can I set frequency & read distance in ARINC 429?
For example can you give an example to tune a 1105.5 MHz and read 25786 ft.
LABEL | INFORMATION CONTAINED IN LABEL | FORMAT | SPEED
------+--------------------------------+--------+----------
035   | DME Frequency                  | BCD    | Low Speed
202   | DME Distance                   | BNR    | Low Speed

Your formatting was pretty confusing.  I took a guess at fixing it up.

In what context? ARINC 429 is just a protocol, and its content is very equipment-dependent. If you have two units on the same bus, usually the SDI bits select one; otherwise some of the data bits may select a channel. You need to read documentation on your particular unit.

The first thing you need to understand is that DMEs are not tuned by their actual frequency but by the frequency of the paired VOR/VHF Com, ILS, or MLS frequency.  For the vast majority of operations, this means a VOR or ILS frequency. This frequency as well as mode settings are coded in the Label 035 as shown below:

And the text from reference to Section 4.1.2:

So you wouldn't attempt to tune 1105.5 MHz. First, it's not a valid DME frequency, as DME channels are assigned to whole MHz.  1105 MHz is valid, but would be tuned by the paired frequency of 113.40 or 113.45 MHz (VOR) or 5067.6 or 5067.9 MHz (MLS). You don't need to know the actual DME frequency as the DME knows the pairing and the DME ground stations are charted using the paired frequency. The snapshot above includes an example of coding to different frequencies.
The unit you mention in the comments is a two channel DME.  DMEs are available with up to 5 channels. Tuning of a specific channel is done using bits 11-13 (DIR FREQ 1-5).  Labels are sent sequentially. The DME expects a minimum of 5 labels per second and any one channel label must be received at least once per second.
The DME distance is output on label 202 as an unsigned twos complement fraction as shown in the following from ARINC 709:

There are 16 significant bits. Bit 28 is equal to 256 NM and each successive bit is half the value the previous. (e.g., bit 27 = 128 NM) The LSB is Bit 14 and is approximate 0.008 NM.
The distance and frequency labels are output from the unit in blocks so you can correlate distances to stations.

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text:
What is the difference between active tip clearance control system and passive tip clearance control system in turbine?
What are the main differences between turbine active tip clearance control system and passive tip clearance control system?

Which tip are you asking about? Tip of the nose, tail, wings, vertical stabilizer, all over them, or am I misunderstanding the question?

@Terry according to the tags it looks like engine fan blade tips. But that’s also just a guess. @ OP: please clarify your question

@Florian I should learn to look at the tags! LOL

sorry guys, I have just edited my question

This is kind of half an answer, but the difference between active tip clearance control and passive tip clearance control is in the definition of “active” and “passive”. An active system has some way of determining the clearance, and then making adjustments to achieve a desired gap. So, it has a feedback loop in the control system. A passive system has no feedback, it just operates “blind”.
Turbine blade tip clearance is usually achieved by blowing cool or hot air onto the turbine casing, to make it contract, and move closer to the blade tips (when cool air is being used), or expand (when using hot air). A passive system will just blow x amount of air at temperature y, during a given portion of the mission (say the climb), without actually knowing what the actual clearance is.  It was up to the designers to make sure the values for x and y result in the right tip clearance. In comparison, l expect an active system has some feedback so it modulates the flow to achieve a desired clearance.
Here is a scientific paper that describes a concept for active tip clearance. The paper uses the term “open loop” to describe a passive system. An active system would be “closed loop”, because there is a signal going back to the controller, telling it how well it is achieving its target tip clearance. It appears in this paper, this signal is not an actual measure of the tip clearance, but rather a prediction from a thermal model of the casing & blade.

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text:
How to decode NOTAM?
I am not a professional, but an amateur trying to self-learn NOTAM code. I need some help in decoding this real NOTAM:

A1671/23 - VIDF (India): Temporary restricted area Activated FIR
Period: 2023-08-25T00:00:00.000Z - 2023-08-25T23:59:00.000Z
Q-Code (RTCA): Navigation WarningsAirspace restrictionsTemporary
restricted areaActivatedChanges
AIRSPACE USE PLAN (AUP) TSA/TRA /AMC-MANAGEABLE AIRSPACE ACT AS FLW :
NR AIRSPACE ID TIME FM TIME TO LOWER FL UPPER FL 1 TRA503 (A) (PUNE)
0230 1030 FL325 FL430 2 TRA503 (B) (PUNE) 0230 1030 FL325 FL430 3
TRA507 (RATNAGIRI) 0330 0800 GND A065 4 TRA508(A) (SHIKRA) 0200 1200
GND A007 5 TRA508(A) (SHIKRA) 1330 1730 GND A007 6 TRA508(B) (SHIKRA)
0200 1200 GND A015 7 TRA508(B) (SHIKRA) 1330 1730 GND A015 8 TRA508(C)
(SHIKRA) 0200 1200 GND A030 9 TRA508(C) (SHIKRA) 1330 1730 GND A030 10
TSA509 (NAVAL DOCKYARD) 0200 1200 GND A015 11 TSA509 (NAVAL DOCKYARD)
1330 1730 GND A015 12 TSA701 (A) (ADAMPUR) 0030 1929 FL145 FL280 13
TSA701 (B) (ADAMPUR) 0030 1929 GND FL280 14 TRA702 (A) (SRINAGAR) 0130
1130 FL170 FL310 15 TRA702 (A) (SRINAGAR) 1400 1930 FL170 FL310 16
TRA702 (B) (SRINAGAR) 0100 1030 GND FL310 17 TRA705 (AMBALA) 0500 0800
FL325 FL500 18 TRA706 (A) (GWALIOR) 0230 1100 FL325 FL460 19 TRA706
(B) (GWALIOR) 0230 1100 FL325 FL460 20 TRA706 (C) (GWALIOR) 0230 1100
FL325 FL460 21 TRA711 (A) (AGRA) 0230 1100 FL325 FL460 22 TRA711 (B)
(AGRA) 0230 1100 FL325 FL460 23 TRA716 (A) (BAREILLY) 0230 1000 FL325
FL460 24 TRA716 (B) (BAREILLY) 0230 1000 FL325 FL460 25 TRA716 (C)
(BAREILLY) 0230 1000 FL325 FL460 26 TSA722 (GRASS BIR - 13) 0330 1130
GND FL280 PART 1 OF 6 CREATED: 24 Aug 2023 11:33:00 SOURCE: VIDPYNYX

I understand some commonly known abbreviations but what would this mean in plain English?

Cross-posted here: https://www.reddit.com/r/flying/comments/160psg8/need_help_decoding_this_notam_to_plain_english/

Yeah trying to find help there too

Why don't you share with us the parts you already understand so someone doesn't have to spend more time than necessary telling you things you already know.

Surely there are on-line decoders-- then you could see the whole thing in plain English and your questions would be answered.  I googled "Decode NOTAMS" and got a lot of hits.  https://www.notams.faa.gov/downloads/contractions.pdf may be what you need.  See also https://www.drorpilot.com/English/notam.htm , https://www.thinkaviation.net/notams-decoded/ , https://www.theairlinepilots.com/flightplanningforairlinepilots/notamdecode.php , and many more.  Feel free anyone to make an answer based on this.

How to decode NOTAM?

Actually this NOTAM is not coded.
The coded version is only used for transfer on the aeronautical fixed service (AFS), a sort of telex network used by aviation entities all over the world. The coded uses only short codes, e.g. temporary restricted area is QRT.
While it is the plain version, it still uses a lot of identifiers and abbreviations. Let's first format the text into a more readable form:
NOTAM A1671/23
VIDF
Temporary restricted area
Activated
FIR
Period: 2023-08-25T00:00:00.000Z - 2023-08-25T23:59:00.000Z

Q-Code (RTCA):
  Navigation
  Warnings
  Airspace restrictions
  Temporary restricted area
  Activated
  Changes

AIRSPACE USE PLAN (AUP)
TSA/TRA /AMC-MANAGEABLE AIRSPACE ACT AS FLW :

NR AIRSPACE ID        TIME FM TIME TO LOWER FL UPPER FL
1  TRA503 (A) (PUNE)  0230    1030    FL325    FL430
2  TRA503 (B) (PUNE)  0230    1030    FL325    FL430
3  TRA507 (RATNAGIRI) 0330    0800    GND      A065
4  TRA508(A) (SHIKRA) 0200

...

PART 1 OF 6
CREATED: 24 Aug 2023 11:33:00
SOURCE: VIDPYNYX

Explanations
NOTAM A1671/23
VIDF
Temporary restricted area
Activated
FIR
Period: 2023-08-25T00:00:00.000Z - 2023-08-25T23:59:00.000Z

This is a NOTAM from the Delhi FIR, related to the activation of temporary segregated areas or temporary reserved areas (which are in Mumbaï FIR). This is usually to give a specific user the exclusive right to fly in a certain airspace at a certain time. The NOTAM is valid for August 25, 2023.
Q-Code (RTCA):
  Navigation
  Warnings
  Airspace restrictions
  Temporary restricted area
  Activated
  Changes

These are categories pertaining to the NOTAM content. They were received as codes, according to RTCA Q-Codes standard. The meaning is straightforward.
AIRSPACE USE PLAN (AUP)
TSA/TRA /AMC-MANAGEABLE AIRSPACE ACT AS FLW :

This is about airspace use (AUP), specifically about manageable airspace. India has establish airspace management cells (AMC). The procedures used are described in the AIP in Manual of flexible use of airspace.
The list of areas concerned by this NOTAM is then given, with an indication of the time widows and altitudes which are regulated, e.g.
NR  AIRSPACE ID         TIME FM  TIME TO  LOWER FL  UPPER FL
1   TRA503 (A) (PUNE)   0230     1030     FL325     FL430

is related to TRA503 (A) which ATC unit is Pune ACC in Mumbai FIR. This reserved area is defined in the AIP, and the controlling authority is the Indian Air Force. This means the military have control of this area, but it can be used under certain conditions by civil flights, and Pune ACC has a delegation for these flights (in practical when the area is active, a civil and a military controller work together).
Note it is said in the AIP "Activation of TRA will be notified through AUP/UUP by AMC", and this NOTAM, which is an AUP, is the required notification.
TRA503 (A) is reserved to the controlling authority between 02:30 and 10:30 (on 25 August 2023), between FL325 and FL430.
The other rows are similar.
SOURCE: VIDPYNYX

This NOTAM was sent on the aeronautical network by the International NOTAM Office of the Airports Authority of India, which code on the network is VIDPYNYX.

Overall: The NOTAM is the notification for the use of certain areas, which are controlled by the air force, but open to civilians. This shared airspace is regulated, and details are found in the AIP as well as the way they are activated, meaning their access can be restricted. These areas are duly shown on aeronautical maps.
More:

How to find out if Restricted Area is "hot" before taking off?
What are some examples of warning areas and restricted areas?
Can a pilot fly through a restricted area while flying on a Victor Airway?
Restricted Area "popping up" around you -- can it happen, and what do you do about it?

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text:
What is preventing a wingsuit with rotors from being made?
This is probably the wrong place to ask, and if so, I apologize, but I was wondering, what are the roadblocks to making a powered wingsuit?
If we look at something like the Vulture from the MCU, is the problem the power source, material weight, or wingspan? Or is it a combination of all three? If we had a suitable power source, could we feasibly construct a similar mechanism?
Again, sorry if this is the wrong place, but I was looking for more realistic answers for a novel I'm working on.

Gravity Industries has made a small-scale version with a few small jet engines.. you can check that out. And for more mythical questions use World Building SE

What specifications are you working from, that you think this HAS NOT already been done?

When does a powered wingsuit stop being a wingsuit?

@DJClayworth, when the body is no longer structural, and flight control is performed by manipulating control surfaces vs twisting and shifting the body.

Closely related: https://aviation.stackexchange.com/questions/91555/are-there-retractable-compact-hang-gliders/91557#91557

Well, they already have. Yvette Rossy developed a jet powered wingsuit a number of years ago.
The chief downside to this design is the inability to land without a recovery parachute on the user.  VTOL models are, apparently, not yet available.
Jet packs and rocket belts have been around for the past 60 years in one former or another, and typically limited by fuel supply and control difficulties in flight.  Gravity industries is coming very close to a product for mass market using micro-sized jet engines for propulsion.  Another example are electrically powered quad copter platforms, which allow for a single pilot to stand on.  I’d venture a guess that the chief difficulty in all of these products is chiefly safety, because it’s incredibly easy to lose control of such a device and either seriously injure or kill yourself.


text:
Requesting to over fly a city “City Tour”
I’m an instrument rated pilot and I plan on taking my family flying for the first time soon. I fly out of FCM airport underlying the Minneapolis Bravo and I wanted to take my family on a flight over the city of Minneapolis. I’ve heard other pilots request a “city tour” while flying before and I didn’t want to go into it confused so how would I approach requesting clearance to overfly the city(which falls into the Class B)?

In my experience there are a few things you should do:

Talk with a local, experienced CFI to find out what the typical course of the "city tour" is. When I was flying out of KPNE near Philly they had a similar clearance that was basically a loop over downtown avoiding some key KPNE points. I was shown the course by my CFI and subsequently flew it with others. Most local pilots knew the course.
Explain to your passengers the tour is clearance based and you may or may not get cleared in the air. This avoids them bothering you mid flight if you don't get clearance
Take off and request a bravo clearance with a "city tour", you will likely get an "affirmative cleared for the city tour" from ATC. You may also get specific instructions like "stay north of XXXXX or west of XXXX" etc.
For this kind of operation, VFR, load permitting, I'd get flight following as well.

Note: You can always start the tour well outside the Bravo, get on with approach for flight following and try to coordinate with them. You will also find it's easier to do these things early in the morning or later at night when the Bravo airspace tends to be less busy.

Since you're IFR rated you can also file an IFR flight plan with direct routing over the city and try to get cleared for it. This, of course, is not a guarantee but for IFR pilots is a possibility. I fly in the NYC area a lot now and when it's busy I generally get vectored around the Bravo but if I file at an off hour it's more than likely I'll get to go through it with pretty decent city views.

congrats on breaking the 100K rep mark!

@DeltaLima Thanks!!!!


text:
How Can I Know If An Airport/ Airfield Is Equipped To Operate During A Power Outage
My buddy was asked this by an IP. Is there a way to see in any publication that an airport is equipped to operate during a power outage? Or a way to see if a airfield has backup power?

Please include the location or legal jurisdiction and avoid less common or regional acronyms (such as IP) in questions.

IP is Internet Protocol. Also IP MAN is great at Kang Fu. Maybe IP means Instructor Pilot who knows.

For lighting at least, it will be listed in the AIP:

You can pretty much assume all aerodromes have backup power for stuff like ATC-equipment etc.

Awesome, thank you for your time. It makes sense that basically everywhere is equipped with some sort of backup power.


text:
What effect does fuel in ACT 1 and ACT 2 have on Cargo Carrying capacity on an A321 aircraft?
Are there any limitations on how much cargo can be carried in H3,H4, and bulk on an A321 if there is fuel in ACT1 and ACT2?
What is the reference manual for this information?

Yes, there are limitations. Go to Weight and Balance Manual - Limitations - AFT Cargo Hold.


text:
How is a "turning polar" constructed?
To start with something familiar: here's a flight polar. As we learned, the best glide speed is found by starting at the 0,0 point, and drawing a line tangent to the sink rate line. The intersection point gives the airspeed for the best glide. In headwind or sink, the polar moves and so does the tangent line, producing different best glide airspeed.

And here is a "turning polar" (from BGA webinar "Thermal Centering"). How is it constructed?

It looks to me that, to draw each small polar (for 20°, 30°, 40°, 50°, 60° bank), the bank angle is fixed and all possible pitch angles are "flown"; the resulting turn radius and sinking speed produce the point on the plot.
But how is that "optimum speed" found, on each small polar (e.g. 43kts for 30° bank)? This will probably also answer the question why does the overall plot of the optimum speed (dotted line) looks what it looks.

I think it's much more intuitive to draw this "turn polar" in terms of airspeed instead of turn radius. Note that bank angle directly relates to centripetal acceleration, which together with airspeed directly results in a turn radius. So instead of "all possible pitch angles", the diagram is constructed using all possible airspeeds.
Optimum speed in this context is minimum altitude loss per unit of angle. Turn rate is equal to centripetal acceleration divided by airspeed, so inversely proportional to turn radius. So instead of a straight line tangent, you draw a hyperbola that just touches your polar to get the optimum speed.
I assume that the fact that a single hyperbola touches all bank angle polars is due to the same reason that a single tangent line touches all different mass lines in a normal polar: a different load factor simply produces the same result at a different air speed. This is probably why they plotted turn radius instead of airspeed. I can't at the moment give a difinitive reason however.

But the dashed line is not touching all the bank angles curves at the point corresponding to the same angle-of-attack (and therefore the same L/D, if we ignore the additional nuance explored in the final paragraph of my answer.). It's not clear to me whether or not this answer is consistent with this observation.

@quietflyer agreed. Some more thinking is required... :)

It looks to me that, to draw each small polar (for 20°, 30°, 40°, 50°,
60° bank), the bank angle is fixed and all possible pitch angles are
"flown"; the resulting turn radius and sinking speed produce the point
on the plot.

Correct, except for the "all possible" part.  Note that the right-hand end of each of the bank angle curves corresponds to the angle-of-attack that would give the minimum sink rate with the wings level.  Moving further left on each of the bank angle curves, we are increasing the angle-of-attack (moving closer to the stall angle-of-attack), which increases the sink rate.  So we are only looking at a very small part of the glider's overall flight envelope in terms of the possible angles-of-attack, compared to the glide polar shown in the upper figure in the question.

But how is that "optimum speed" found, on each small polar (e.g. 43kts
for 30° bank)?

Imagine that rather than drawing only curves for 5 bank angles between 20 and 60 degrees, curves were drawn for every possible bank angle in between.  For any given turn radius, the dashed line is drawn through whatever airspeed on whatever bank angle curve gives the lowest possible sink rate.
Note that the diagram indicates that at steep bank angles-- which result in high airspeeds due to the high wing loading-- there is an advantage to flying at an angle-of-attack higher (closer to stall) than the one that would generate the minimum sink rate for that bank angle, if the end goal is to minimize the sink rate for a given turn radius.1
It would be interesting to know whether the curves for the various bank angles were generated by simply "adjusting" the wings-level glide polar for the increased wing loading, or by actual measurements.  In turning flight, the "relative wind" is actually curved in all three dimensions-- for example, a string trailing from the tail would stream in a different direction than a string attached to the front of a probe projecting forward from the nose, due in part to the continual nose-up, tail-down pitch rotation of the aircraft in the turn.2 This has aerodynamic consequences, such as an apparent decrease in the decalage angle between the wing and tail, requiring more up-elevator (or less down-elevator) to maintain any given angle-of-attack when the aircraft is steeply banked and turning tightly, than when the wings are level.  These aerodynamic changes associated with turning flight affect drag, and sink rate.3 Therefore the L/D ratio does vary to some degree as the bank angle is increased and the angle-of-attack of the wing is held constant.  These aerodynamic consequences associated with turning flight are most pronounced in slow-flying aircraft, where the radius of the turn is not orders of magnitude larger than the physical dimensions of the aircraft.
Footnotes:

The reason for this is that if the glider were constrained to fly at the min-sink-rate angle of attack rather than a higher angle-of-attack, a higher bank angle would be required to achieve a given turn radius.

This is a bit tricky-- we're really trying to describe the direction of the undisturbed relative wind, referenced to various different points on the aircraft.  We're not including the downwash from the wing, or any other effects caused by the physical interaction between the aircraft and the airflow.  This change in direction of the undisturbed relative wind also causes changes in the direction of the actual airflow experienced at various points on the aircraft.

The ideal fuselage shape for minimizing drag in thermalling flight would be bent like a banana, convex side down, curved to conform to the curving flight path and relative wind.

The polars in the question look like they have been copy-pasted from a straight flight polar and fit so the results look OK. In truth, the steeper polars also look more compressed in X direction.

@PeterKämpf - interesting observation--


text:
What is the nature of the glider's vario lag?
I learned that glider's vario reading lags behind the vertical speed of the airmass. For example, when flying a steady circle and having a thermal not quite centered yet, the vario is reading a maximum well after the core has been left. See BGA Webinar "Thermal Centring".
The implications for thermal centering are explained e.g. in the BGA Instructor Manual Ch. 24. "The average mechanical variometer is about three seconds too late".  .
What is the nature of this lag? Wouldn't a gentle blow into the static port make the needle move immediately (certainly without a delay of a few seconds between strongest blow and strongest needle deflection)?

Has it to do with the the traditional vario's capacity flask and capillary hole or does it apply to newer electric variometers operating on the transducer principle as well? (A tiny vacuum cavity on a circuit board is sealed with a flexible membrane with embedded strin gauges.) See FAA-H-8083-13A, Glider Flying Handbook, Fig 4.21
Or is it the glider's mass (inertia)? A glider can't change its vertical speed instantly.

I'd always thought of vario lag as being around 1-2 sec.  The whole point about them vs conventional VSIs is they are nearly instantaneous.  3-5 secs is more like a conventional VSI.

@JohnK -- I've always understood-- must have read it somewhere-- that conventional VSI's were intentionally constructed to average the vertical speed over a period of about 7 seconds.

The vario is essentially a high-sensitivity vertical speed indicator, operating on the same principle: the air flow rate in or out through the orifice is measured as a pressure difference between the inside of the instrument and the outside of the instrument.
Hence, the sensing apparatus responds as a resistance (the orifice) in series with a capacitor (the capacity flask) and this mathematically yields a system response time constant which is the root cause of the lag.
The time response of a vertical speed indicator can be improved by adding an "accelerator pump" in the form of a small spring-loaded piston that can rise or fall in its cylinder in the vertical axis. The cylinder is connected to the inside of the instrument case so if the plane suddenly rises, the piston on its spring slides down and pumps a little air into the case and thus forces the gauge needle to get with the program quicker. The opposite happens when the plane suddenly descends: the piston on the spring rises and pulls a little air out of the case and urges the needle to more quickly indicate a descent.
The sprung piston only responds to sudden movements; for slow movements it stays out of the picture.

OK, so re content in question "Has it to do with the the traditional vario's capacity flask and capillary hole or does it apply to newer electric variometers operating on the transducer principle as well?", you are saying "yes" to the former and "no" to the latter?  (That would tend to be my opinion, although my flaskless Brauniger IQ Comp hang gliding variometer still has at least .5 seconds of lag I believe-- )

And how much role do you think this plays into it: "Or is it the glider's mass (inertia)? A glider can't change its vertical speed instantly."?  Which is honestly a point that I had never given much thought to before--

Re " although my flaskless Brauniger IQ Comp hang gliding variometer still has at least .5 seconds of lag I believe-" -- this observation is based on holding the vario in my hand and abruptly moving it upwards as far as my arm can reach, so the inertia of the aircraft in coming into equilibrium with the updraft plays no role in this observation!

One more observation-- nowadays, of course, the most advanced variometers for sailplanes (and now for hang gliders/ paragliders too) use peizoelectric accelerometers, not mechanical "accelerator pumps".  Which naturally has led us down the path to full AHRS systems incorporated into modern sailplane variometers, so that one or more of the display screens includes an artificial horizon display--

I wonder if there is a feasible way to find out or to measure the lag. Motivation: misjudging the lag by two seconds leads to performing straightening up & turning again about thirty degrees away from what would be the best point of the circle for doing this.


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Use of Fuses in modern Aircraft Systems
Are Fuses allowed in the control electronics of an airplane?
Are Fuses allowed in the electronic power paths which supply actuators?
Thank you in advance.

Are you asking specifically about fusible links or over-current circuit protection including circuit breakers too?

No, Jim. Until now I didn't use fuses in the projects I followed. I did so only because i was told it wasn't a good practise and it wasn't allowed. Unfortunately I didn't find this rule written anywhere and here it is my doubt.

Yes you will find fuses sometimes used where you need instantaneous protection of a circuit from overcurrent, where a circuit breaker, which must heat up to work like the fuse does but needs more time to do so, takes too long to function.
Normally you will see them protecting sensitive electronics where even a short overcurrent state can blow up a $50,000 circuit card within a black box, and the fuse may be built into the box itself.
You can use a fuse in place of a circuit breaker anywhere you want, since they are considered more reliable and faster acting than circuit breakers, but you have practical considerations with fuses, like the need for spares and the difficulty in resetting the circuit, and the fact that in some (or most) circuits, the transient overcurrent tolerance of a circuit breaker is desirable.
So you'll use a fuse where a fuse's technical characteristics are really necessary and aren't available by other means, but otherwise a circuit breaker is a better choice.

Thank you John K for your answer. I agree on what you said. Do you know if DO-254 talks about that?

There’s another use case for fuses, that are quite common in GA aircraft.  For instance, in many Cessnas, the Hobbs meter and clock are protected by a fuse, not a breaker, because you don’t want people to be able to turn them off.  The battery contactor (what most people would just call a relay) for ground power is also fused.  I would add that circuits that are prone to arcing, like a ground power connector, may be better suited for fuses.  Circuit breakers will be more resistant to transient voltages, but may not handle an arcing situation correctly.


text:
Textured Spray Paint for Promoting Turbulent Flow
I'm looking for textured spray paint to apply on my RC model to ensure turbulent flow on a few of the smaller aero surfaces.
Does anyone have any experience doing this and have any recommendations on specific spray paints I could use? Also open to any other coatings as well
Thanks

Do you know what surface roughness you are aiming to achieve?

Related: https://aviation.stackexchange.com/q/67733/19

Do this. It's how large no-skid walking zones are made on airplanes, boats, etc:

Mask off the area you want textured.

Apply a coat of regular spray enamel paint. Whatever you have lying around, or a specific colour to make the modified area stick out.

While the paint is wet, sprinkle clean sand into the wet paint. Try a salt shaker to get an even distribution of the sand particles. For different surface roughness levels, just use different size sand particles, and apply more or less density.  Experiment on a scrap piece of material to get the effect you want.

Apply another coat or two of paint over the sand.

If you don't like the result, just remove the whole coating with some Varsol before it fully dries, and try again.

Fantastic idea - thank you!

This is also how industrial floors are often treated if high friction is desirable. Sand is sprinkled over fresh pain. Very effective.

Instead of paint, I would suggest you get some anti slip tape and apply that to the surfaces. The main benefit of the tape is that you can easily modify the area according to results you get from testing.
There are many options availlable, but I think they are all pretty similar in grit, maybe P40 to P80.
I'm not aware of textured spray paint that would do what you are looking for. While some leave an uneven or crackled surface, they still are pretty smooth and glossy, and the texture will not be coarse enough to effectively separate the flow.

FYI, there are stone textures that are fairly rough:  https://www.rustoleum.com/product-catalog/consumer-brands/american-accents/stone-spray

Excellent and simple idea - thank you!


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Impact to lift curve slope for a tapered wing with only trailing edge sweep?
I'd like to use a tapered wing with 0 deg leading edge sweep and 12 deg trailing edge sweep as shown below for my RC model. I understand that lift curve slope, a, is reduced for wings with leading-edge sweep since the normal velocity component is lower than the freestream velocity, resulting in a lower pressure difference between the upper and lower wing surface.
Would there be a reduction in lift curve slope or any other performance (compared to a straight wing, with no sweep), for my wing with only sweep at the trailing edge?

Since we are talking about an RC model, the different Reynolds number between tip and root might have a not negligible impact. Plus all what @Koyovis has already explained.

Would there be a reduction in lift curve slope or any other performance (compared to a straight wing, with no sweep), for my wing with only sweep at the trailing edge?

1. Lift curve slope. Leading/trailing edge edge sweeps of 0° & 12° yield a quarter chord sweep of 3°, which is very little. When for instance using the simple sweep theory like in this answer, wing lift coefficient $C_L = c_l \cdot$(cos$\Lambda)^2$ reduces by (cos 3°)$^2$ = 0.997 = 0.3%.
2. Any other performance. Stall characteristics might be slightly improved, since the trailing edge will deflect the flow slightly in the wing root direction, so full aileron control in a stall. You might find more info in the answers to this question, although they pertain more to faster planes with higher sweep angles.
3. Structural implications.. The usual issues with forward sweep are increased tip torque and increased flutter tendencies, which requires stiffer torsional construction. But again, with a straight leading edge and only 3° forward sweep there should be no serious consequences.

Above picture (from Wikipedia) shows the L-13 Blaník. Quoting the Wiki article, it is the most numerous and widely used glider in the world. It has slight forward sweep for positioning the instructor not above the main wing spar, and has very favourable handling characteristics.


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Spanwise location of mean aerodynamic chord for Swept Wing
How would I calculate the spanwise location of the mean aerodynamic chord for a swept wing?
I see a general formula for an elliptical planform wing, but having trouble finding anything on swept wings.

Formula for Elliptal WIng:

Basically I think you do a weighted average based on the wing chord width. Will let someone with more expertise post the actual answer. Interested to see if anyone has anything to say about how to take twist (washout) into account. For example, what if the outer fifth or quarter of wing is so washed out as to be negatively lifting at moderately fast airspeed (moderately low AoAs)?  I've seen this happen with hang gliders, and not primarily due to the wing flexing.   In such a case do we say the MAC moves inboard (and thus forward) as the AoA decreases? Maybe grounds for a new question...

Take a look at https://skybrary.aero/articles/mean-aerodynamic-chord-mac#:~:text=Definition,of%20a%20tapered%2C%20swept%20wing.  Take it with a grain of salt though, for example I think "measured parallel to the normal airflow over the wing" is wrong.  You measure parallel to the aircraft centerline.  I assume they are just giving the formula for an average, weighted according to chord length.  In the case of linear leading edges and trailing edges it should be fairly simple.

For a conventional trapezoidal wing the spanwise location of the mean aerodynamic chord has a simple equation and it can even be determined geometrically. The following picture (source - The original source of this picture is anyway the book Airplane design by Daniel P. Raymer) presents both methods ($\lambda$ is the taper ratio):

Perfect, just what I was looking for - thank you!

The spanwise location of the MAC is a meaningless thing.
Even the length of the MAC is fairly dubious.
The thing that matters is the location of the aerodynamic center (of a wing or a wing/tail combination).
The length of the MAC is used as a reference chord -- to non-dimensionalize moments.  You can use any value so long as you are consistent.  You can use the MGC (mean geometric chord) and everything will work out fine.  Perhaps there is some empirical data out there that is fit to MGC that won't quite match your MGC use, but that isn't the end of the world.
We would typically like the c/4 of the MAC to align with the wing -- and even better if that is at the location where the wing's chord is equal to the MAC.  However, there is nothing that defines that to be the case.
When I say that the spanwise location of the MAC is meaningless -- I mean that in terms of "What do you actually use it for".  Perhaps I'm overlooking something, but I can't think of a use of that term in either roll or yaw calculations.  I.e. it is something that looks important from the often geometric construction of the MAC -- but it doesn't actually matter.
Happy to be proven wrong on this.

I'm using it to calculate moments for a novel aero body with a swept tail surface. So I ultimately need the moment arm distance from a specific point to the cp of the tail. In my case, the cp would be at 25% of the mac (since it's a symmetric airfoil). Knowing the spanwise location of MAC would allow me to calculate the chordwise location from the root leading edge using trig.

I do not believe most methods of deriving the MAC will yield the spanwise position -- and if it did, there is no requirement that the MAC lie 'on' the wing.  It can lie in front of or behind the wing.  I suggest you use a LVM code like VSPAERO, AVL, XFLR5, Tornado, etc. to perform a simple pitch stability calculation for your configuration.


text:
Has the use of a radiator surface on the inside of duct ever been tried?
I wonder if using radiating surface on the inside of a ducted fan to dispel the waste heat of an engine would provide an economic advantage by increasing thrust?  Since a typical engine “wastes” some 40% of its energy as heat could it not be used to expand the air going through the duct a little like a jet?

That's basically what happens with the radiator on the P-51.  It uses the Meredith effect to use the heat from the radiator to provide a small amount of thrust.


text:
Has anyone used AI to design airplanes?
Photo example
Prototype drawings
Any response from someone who has done it.

Would you mind using complete sentences in your question?  A list of bullet points is fine when introduced with something that a grade-school teacher would recognize as a sentence: a subject, a verb, appropriate punctuation, etc.  As it stands, this question asks for quite a bit of information in response, but shows the very slightest possible effort at presenting what it is you're asking for.

AI is an extremely broad definition encompassing hundreds (literally!) of methodologies.
For sure some of them have been used in the aerospace world: genetic algorithms, Fuzzy logic, speech recognition, decision tree, image recognition, optimisation algorithms...
If you follow this advice maybe we can narrow down the scope and give one more precise and useful answer.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

The question is if "Has anyone used AI to design airplanes" My answer is yes and I gave some examples. Plus, I'd be glad to give a more specific answer if requested. Why that shouldn't answer the question?

Aviation Youtuber Bryan Turner has done exactly this.

-1  What happens to your answer when/if this video is removed from YouTube?


text:
Calculate/Measure Sideslip Angle from Heading Track and Drift Angle
Is there a way to measure the sideslip angle during flight through the relationship of heading, track and drift angle? For example if you turn the aircraft into head or tailwind - does the sideslip angle equal the drift angle for establishing a steady heading sideslip through rudder and aileron input from this initial condition?

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

It's a little late now as several answers have been posted (including mine), but I'm realizing that I should have asked for a clarification of exactly what you mean by the "drift angle".

Sideslip has nothing to do with wind, or drift. No matter what the wind is, unless you apply rudder or some other yaw inducing force, (Single engine p-factor or differential thrust on a multiengine aircraft), the aircraft will always weathervane into the relative wind, the ball will remain centered, and there will be zero sideslip.

I would argue, that this does not answer the question. The question is how do you identify sideslip given these inputs, not how does sideslip work in motorized aicraft. You sentence is not true for example for rotorcraft such as helicopters or drones

I thought that saying that it has nothing to do with wind or drift implies the answer to the question..... It is a resounding, unequivocal NO! You can't use wind, heading, drift angle or anything like that to measure sideslip. The only way to measure sideslip (Helicopter or fixed wing aircraft) is with a ball or a yaw string...

... and the sentence is true for helicopters!  ... and drones. Unless there is some rotational torque on a Helicopter, like the torque from a single rotor, (Not a Chinook), or the torque from a tail rotor like on a conventional single rotor helicopter, a Helicopter will weathervane into the relative wind just like any other body free to move in a fluid.

You are right that the steady-state response will go to zero for all weathervaning aircraft, but the dynamic response is very much non-zero due to factors like transient wind or wind shear and due to mismatches in torque as a result of the extensive coupling in helicopters. This last point can be very nicely seen when looking at logged Beta data (I just took a look to make sure). Steady state is practically never reached, and the amplitude can be as high as 15° either side. Additionally, a lot of drones tend to not weathervane, e.g. the typical Quadrotor.

No.
Drift angle is just the difference between your heading and your ground track.  There is no way to know whether an aircraft's heading is different from its ground track due to wind, or due to full rudder input by the pilot.
More information would be needed to determine whether or not there is any sideslip angle.

Yes, in theory it would work to establish the sideslip angle from the difference in heading and track angle, if you are traveling exactly into the wind or downwind. In practice, the challenge will be to exactly determine the wind velocity, and making sure your track is exactly aligned.
The answer will depend on the definition of drift. So before answer, let me introduce my definitions of these terms, so it is clear what I am talking about.

heading: the angle between the north (true or magnetic) and the x-axis of the body frame of reference of the aircraft. The x-axis is aligned with the
fuselage.
side slip: the angle between the incoming air and the plane of symmetry of the aircraft.
drift1: the angle between the motion of air, relative to the aircraft and the motion of the ground, relative to the aircraft. Ignoring tectonic plate movement, drift is caused solely as an effect of  wind.

A picture says more than a thousand words, they say:

So, if you align the track angle (relative ground motion) with the relative air motion (0 drift), and at the same time introduce a sideslip, the difference between the heading and the track, is purely caused by sideslip.
In practice, that will be extremely difficult to achieve that in a steady state.
When the track angle is not aligned with the relative wind motion, you can compute the side slip angle if you happen to know the exact magnitude and direction of the wind.
It is more practical to measure the sideslip angle by using a beta vane on the aircraft.

1
An alternative definition of drift is: drift the angle  between the x-axis of the body frame of reference of the aircraft the motion of the ground, relative to the aircraft.
In that definition, drift is the combined result of sideslip and wind.

Terminology makes for such good internet discussion!

Calculate/measure sideslip angle

That means one plans to fly uncoordinated, using a sideslip to hold heading.  While this technique works to keep the nose centered on the runway when landing, it is not the best way to fly a crosswind over distance.
What is more efficient is to fly a coordinated heading and let the wind "correct" the groundtrack.

the coordinated airplane has no crosswind relative to the airmass, the vector sum of aircraft velocity/direction and windspeed/direction determine groundtrack.

So, ground track is based on these two components, not the orientation of the aircraft.  That's what this helps a pilot/navigator do.
Regarding calculation of bank angle vs crosswind in a slipping condition, side force from wing is readily available by decomposing the lift vector as sine bank angle × Lift.  But one should keep in mind that crosswind forces are created by drag, essentially the wind hitting the side of the plane.
It would be easier to measure bank angle vs crosswind for a given aircraft, and one could expand the study to crab configuration, power on and off (I can see the graphs on slides and overhead projector already!), but, as wind varies, pilots generally must pilot by constantly adjusting control inputs to the weather conditions, not with data graphs.

Preface:
The external, meteorological wind creates no tendency whatsoever for an aircraft to sideslip.  Nor is sideslip an efficient or logical way to correct for a crosswind, except during the last stages of final approach, when there is a benefit to ensuring that the aircraft's heading is fully aligned with the ground track.  "Crabbing" and "slipping" are two completely different techniques-- "crabbing" is a really no more than the selection of a particular heading to fly, and normally does not involve any sideslip at all.
Here are some related ASE questions whose answers may shed some light on these points:
Is drift and sideslip angle the same in the accompanied figure?
How does sideslip indicator react during crosswind?
Since we usually fly with zero sideslip as best we can, and experience no more particular tendency to slip when flying crosswise to the wind than any other time, your questions are pretty much moot as far as all practical flying and navigation is concerned.
Still, one could imagine a theoretical situation where a navigator was provided with a precise measurement of an aircraft heading, and a precise measurement of a (small but non-zero) sideslip angle, and was interested in computing the exact ground track and drift angle.1
So to address your actual questions:

Is there a way to measure the sideslip angle during flight through the
relationship of heading, track and drift angle?

It depends on what you mean by the "drift angle".
If by "drift angle" you mean the angle between the direction of the flight path relative to the airmass, and the ground track--  i.e. the "crab angle" (or more precisely, the negative of the "crab angle") -- then the answer turns out to be "yes".  See "Calculations part 1" for more.
But if you literally mean the angle that you would measure with a drift meter-- the angle between the aircraft's heading and the ground track-- then the answer turns out to be "no".  See "Calculations part 2" for more.
The reason there is some ambiguity about which of these is the more appropriate definition of the "drift angle", is that sideslip is usually assumed to be zero in wind correction problems.  But you aren't making that assumption, so we need to decide which definition to use.
Likewise, a word about "heading" -- what we usually call "heading" in wind correction problems, is really only heading if sideslip is zero.  If sideslip is not known to be zero, a better term in many cases would be "direction of flight path relative to airmass" (dofprta).  In this answer, I'll use "heading" to mean the actual heading.    I'll also assume that that's the usage that was intended in the question.
Calculations part 1--
Here we define "drift" as the angle between the direction of the flight path relative to airmass (dofprta), and the ground track.  This is simply the negative of the "crab angle".
(Note that by this definition, "drift" is zero at all times when we are flying in still air, even if we are sideslipping.)
Sign conventions used here:
A positive drift angle means that the ground track is clockwise of the dofprta. A positive sideslip angle means that the aircraft nose is pointing clockwise of the dofprta.
"Drift angle" and "crab angle" are essentially the same thing (more precisely, the "drift angle" is the negative of the "crab angle".)
Dofprta + drift angle = ground track
Sideslip angle = heading - dofprta
Therefore sideslip angle = heading - (ground track - drift angle)
Which means that sideslip angle = heading + drift angle - ground track
So yes, with this definition of "drift angle", we can calculate the sideslip angle from the drift angle, ground track, and heading.

For example if you turn the aircraft into head or tailwind - does the
sideslip angle equal the drift angle for establishing a steady heading
sideslip through rudder and aileron input from this initial condition?

It's a little unclear what "turning the aircraft into the wind" means.  Here we'll assume that the questioner is envisioning that we have changed the flight path so that we are flying directly into the wind, before beginning a sideslip.
If we start by flying directly into the headwind (or tailwind), with zero sideslip, then heading, dofprta, and ground track will all be the same, and drift angle (and crab angle) will be zero.  From this point --
a) If we keep the dofprta constant, and establish sideslip by changing to a new heading (while countering with bank to avoid any turn), then ground track will remain constant, and slip angle will equal heading - dofprta (which will also be heading - ground track), and drift angle (and crab angle) will be zero.  (Using these terms as defined above.)  So when we establish a sideslip while keeping the dofprta and the ground track pointing straight into the wind, the sideslip angle does not equal the drift angle.
b) If we keep the heading constant, and establish sideslip by using the controls in such a way that the flight path (dofprta) curves (turns) a bit before stabilizing in a slightly different direction, the situation has gotten considerably more complicated.  Since the flight path (dofprta) now has a crosswind component, so does the ground track.  The drift angle can no longer be zero.  For a given angle between the wind direction and the dofprta, the resulting ground track (and therefore the drift angle) will depend on the relationship between the wind speed and the airspeed.  It will still be valid to say that sideslip angle = heading + drift angle - ground track, which means that drift angle = ground track + sideslip angle - heading.  So as far as case b) is concerned, as to your question "does the sideslip angle equal the drift angle?", the answer is "only in the specific case where the ground track and the heading are the same."  But with the airplane pointing straight into the wind, this can actually never happen unless the sideslip angle and the drift angle (as we've defined it here) are zero.  For all other cases, the answer to your question is "no -- when the airplane is pointing straight into the direction of the external, meteorological wind, the sideslip angle can not equal the drift angle".
In fact, regardless of wind direction or speed, it is always true that if the ground track and the heading are the same, then the sideslip angle must be equal to the drift angle, as we've defined it here, and vice versa.  (There's no reason to ever fly an aircraft in this manner-- except on final approach, using the wing-down method of crosswind correction!)  With a little reflection on the dynamics of wing-down crosswind landings, it becomes immediately obvious that to fly in a straight line with the ground track equal to the heading, with a sideslip angle other than zero, then the following must be true: 1) The ground track and heading cannot be directly aligned with the wind.  2) The direction of the flight path through the airmass (dofprta) cannot be directly aligned with the wind.  3) The dofprta must lie somewhere between the ground track and the wind direction.  These conditions were not met in cases a) or b) above, hence the "no" answer to your question "does the sideslip angle equal the drift angle?" for either of these cases.
Calculations part 2--
Here we define the "drift angle" as the angle that we would measure a drift meter-- the angle between the aircraft's heading and the ground track. (This is arguably the best or at least the most widely-used definition of the "drift angle"-- for example, in the wing-down crosswind landing referenced immediately above, we would normally say our goal is to land with "zero drift", and this concept isn't really fully compatible with definitions used in part 1.)
(Note that by this definition, "drift" is not zero when we sideslip in still air.)
Sign convention used here:
A positive drift angle means that the ground track is clockwise of the heading. A positive sideslip angle means that the aircraft nose is pointing clockwise of the direction of the flight path relative to the airmass (dofprta).
Dofprta + (drift angle + sideslip angle) = ground track
Sideslip angle = heading - dofprta
With these equations, there's no way to state the sideslip angle purely as a function of the drift angle, ground track, and heading.  All we can say is that Ground track = heading + drift angle, regardless of the sideslip angle.  The "drift angle" gives us no new information that we don't already know from the heading and the ground track.
Therefore with this definition of "drift angle", there is no way to calculate the sideslip angle solely from the drift angle, ground track, and heading.

For example if you turn the aircraft into head or tailwind - does the
sideslip angle equal the drift angle for establishing a steady heading
sideslip through rudder and aileron input from this initial condition?

Again it's a little unclear what "turning the aircraft into the wind" means.  Again we'll assume that the questioner is envisioning that we have changed the flight path so that we are flying directly into the wind, before beginning a sideslip.  This introduces additional constraints to the situation.  While flying directly into the wind or directly downwind, if we initiate the sideslip in such a way that the ground track (and therefore the dofprta) remain constant, rather than simply initiating a bank while holding aircraft heading constant, then and only then with this definition of "drift angle", can we say that the sideslip angle is also the drift angle.  (Actually, there is one other case where we can make the same observation: if we know there is zero wind.)
Footnotes:

Here one is tempted to consider the case of a multi-engine aircraft with one engine failed, or with one engine intentionally shut down for enhanced range, but we must remember that in such a case the deflection of the slip-skid ball does not actually indicate the sideslip angle of the aircraft, i.e. the angle between the aircraft heading and the oncoming relative wind-- so how would the sideslip angle be known?  And to the extent that it were known, why would it be allowed to be other than zero?

Your explanation is sound, but the answer to the question, (as it is worded...) is NO.  You cannot compute a sideslip angle from heading, track, and drift angle.

But the question wasn't about "DOFPRTA"...  You state: "Since we usually fly with zero sideslip as best we can, and experience no more particular tendency to slip when flying crosswise to the wind than any other time, your questions are pretty much moot as far as all practical flying and navigation is concerned."  This would seem to bolster my point.

Well, my comment stands.  I think you have way over-answered a pretty simple question, going as far as inventing a new term to justify a possible yes answer…

(Continued in chat https://chat.stackexchange.com/rooms/140720/discussion-between-quiet-flyer-and-michael-hall )

(Sorry, thought there was nothing more to say, but a new answer inspired a little more thought-- realized I hadn't specifically addressed the second sentence in the question, for the alternative (second) definition of drift -- )

Might also be worth noting the impracticality of any technique that requires flying exactly into the wind or exactly downwind... and probably should replace the clunky "dofprta" with something like "air track" or "direction of airspeed vector" -- but answer is telling me it is tired and needs to rest a while --

Deleting comments, replying in the chat


text:
What airfoil can be used in making a turbofan engine fan blade?
The thing is, I am trying to make turbofan blades in Solidworks for a possible undergraduate thesis. I would like to know if there is any airfoil from the NACA series or anything plausible for this application. I would also like some leads on the design process involved in turbofan blades, as this might help if no specific airfoils are available for the public. I'll also conduct CFD analysis on it, and any tips would be appreciated. Thanks!

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Is it better to delete this one now to remove repetition?


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Can current air to air missiles shoot each other down?
Assuming that a missile is able to track, lock and effectively guide itself to another missile for sake of argument, would say an AIM 9X or AIM120 be able to shoot down its respective counterpart?
Note the AIM 9X and AIM 120 are just examples, I'm talking about in general. Air launched air to air missiles are the scope of the question. So anti-ballistic missiles, cruise missiles etc are out of scope.
The reason I ask is because active protection systems against kinetic energy penetrators, especially depleted uranium rounds, face significant challenges in achieving hardkill. An APS system isn't going to do much other than breaking it up a bit or causing the incoming round to tumble or wobble. Either way an APFSDS round is smashing against hull armor and punching a nasty gash at the very least. Speed is one of the issues that make APS systems ineffective compared to stopping HEAT.
Missiles on the other hand are launched way higher in the sky, with a high initial velocity and a powerful rocket motor. While the motor quits eventually, there's still a lot of kinetic energy. However, missiles are more beholden to the laws of aerodynamics and loss of kinetic energy. I'm not entirely sure if blast frag or a continuous rod warhead could expand fast enough to cut up a missile. Or if there was non catastrophic damage, would the damaged missile body and fins allow a missile to still limp its way to a target for a successful kill, or would the missile bleed kinetic energy like crazy eventually falling out or self detonating.

Because the process involves tracking, locking and guiding, your first paragraph essentially asks: Assuming a missile could shoot down another missile, could a missile shoot down another missile?. What specific element of “shooting down” are you NOT assuming, and therefore asking about?

@Michael Hall - That was my question exactly. That whole first paragraph is exactly how you shoot down something else. Acquire, track, guide, close, explode.

@MichaelHall actually a good point :). I simply answered a bit broad...

@MichaelHall The reasons I specified the tracking part was because I wanted to avoid a conversation about if a missile can track another missile. I'm more concerned about if a missiles warhead can detonate in time and yield enough damage to cause a missile to not reach an enemy. A parallel would be me asking if an APS system can destroy a KE penetrator assuming that an APS could track, react, and fire fast enough. In the question here, the focus on is the damage from an APS good enough to counter APFSDS. To which the answer as of now is no because the rod is still flying incredibly fast.

if a missiles warhead can detonate in time and yield enough damage to cause a missile to not reach an enemy You only have one question: can it detonate in time. If the warhead can take down a plane it's more than enough to take down another missile. The hard bit is fusing (just as the hard bit of anti ballistic missiles is fusing)

@KorvinStarmast I'd assume good positioning would equalize things a bit more compared to ABMs. One of the issues w/ ABMs is the reentry speeds, which far exceeds any missile

May not be your best assumption.  The flight path of an IR guided missile is different from the flight path of a beam riding missile is slightly different from a semi active/active homing missile.

Yes they can, although they might have a hard time doing so.
General
Generally speaking, a missile searches for a suitable signal source which it tries to hit, e.g. an IR source such as an engine, or a radar return. Once aquired it alters its trajectory such that it hits the signal source. Therefore for a missile it is no different if the heat source it tracks is the hot turbine engine of another jet, or the (still) hot rocket engine of another missile. The same hold for radar guided missiles.
Example
I think a good example is that of the Diehl IRIS-T Anti-Air missile. Originally conceived as an air-to-air missile, it has been converted to an surface-launched anti-air missile by mainly upgrading the rocket engine, therefore equipping that missile with more energy. This missile was recently deployed to Ukraine in order to defend against incoming missiles. While the missiles the IRIS-T is defending are not typically air-to-air missiles, it shows that a (converted) anti-air missile is capable of hitting another missile.
Physics constraints
However one has to realize that an air-to-air missile is not a good fit to shoot down another air-to-air missile. Both missiles are roughly equally fast and equally maneuverable, however typically you would want your defending missile to have an advantage of higher maneuverability (a value I heard often is 3 times more maneuverable) then your enemy. Additionally, the flight time even of long range A/A missiles is short (at longest a couple of minutes), making targeting and tracking difficult. The angle at which the incoming missile can be tracked and killed is small, the position from which the defending missile has to be launched is narrow. Therefore your defending missile is not as likely to hit the enemy missile
Effector
I am no expert in this, however consider this: Even small holes or damages create huge air-drag at a speed of mach 3. Therefore if you manage to damage the enemy missile even slightly (perhaps a couple of holes, or half a fin), I would say that this missile is doomed.

Interesting. In that case lets say an AIM 9X or a Python 5 is tracking onto an AIM120 or even an AIM54 since the former are far more maneuverable since they have a closer weapons envelope. I'm assuming the annular blast frag from them would shred apart the missile's ability to maintain aerodynamic stability and guide for a high probability of kill?

Iris-T can hit cruise missiles, possibly short range ballistic missiles in boost. Those are quite different targets than an air-air missile (or SAM). Much larger, less agile, and in case of the cruise missile typically slower too.

@FIRES_ICE As I said, I am no expert in damage of those effectors, however from my understanding pretty much any damage to the outside skin or aerodynamic surfaces of an air-to-air missile is fatale to that missile because the high velocity means very little margin for error in terms of aerodynamics

@jwenting Again, killing an air-to-air missile via another air-to-air missile is possible. Such missiles are neither stealthy (90° angles, no stealth optimizations) nor cold (compression heat and hot rocket motor), but the likelihood is bad, as the interception parameters are bad. Both missile are equal in terms of energy and maneuverability. The angle at which the incoming missile can be tracked and killed is small, the position from which the defending missile has to be launched is narrow. Therefore a kill is unlikely, but possible.

"3 times more maneuvrable": how do you define manoeuvrability? Turn rate at maximum speed?

@U_flow Can't an AIM 9X pull 60Gs combined w/ high off boresights capability? I'm surprised that's not enough against something like an AIM 120 or AIM 54. Would all aspect target acquisition improve the odds? As in an AIM 9X will always be guided frontally at an incoming missile, where positioning matters more than KE.
Also, if it sounds like I'm arguing or discounting your answer I apologize for it. I'm just trying to do some research and I've been going in circles. I'm trying to see if a missile can be utilized as either active protection systems or against other AAM reliably.

@Robe The maximum lateral acceleration is normally used in units of a load factor. E.g. an R-73 Wympel can pull around 60Gs, an IRIS-T apparently  in excess of 100Gs.

@FIRES_ICE The intercept angle depends on how the agressor missile is maneuvering and how fast it is. The faster, the narrower the intercept angle will be. If that angle is outside of your tracking angle capability, then you will not be able to track that missile. If target data can be submitted by an external source, it might be a bit different. It is not possible to answer this without looking at specific engagments. Perhaps you can write a (2D) simulation in Matlab or Python to build more intuition. For whatever you are trying to achieve, such a simulations might be helpful.

@Robe Maneuverability is normally defined as the capability of lateral acceleration, wherebey the unit of measurement is the maximum load factor. Therefore, the higher the lateral acceleration is, the tighter the turn can be. E.g. an R-73 Wympel can pull around 60Gs, an IRIS-T apparently in excess of 100Gs. Note that at the speeds of AA missiles, the limiting factor is not how much aerodynamic lateral acceleration can be flown (that is virtually unlimited), but how much your missile can take from a structural point of view. Therefore how many G's can be missile handles before it breaks apart

A more  significant question is how would a pilot launch it? Most A/A missiles have very short Times Of Flight (TOF) - on the order of six to ten seconds. And obviously, the launch window to intercept it would have to be within the early part of its flight. (Intercepting the flaming wreakage after it has hit its target would be pointless!). So How would a pilot acquire, track, re-position his aircraft to satisfy launch parameters, and launch his missile, in such a short period of time?

Also, the maneuvering capabilities of either missile are only relevant if the target is maneuvering to counter the interceptor missile instead of against its target.  If that could happen, (it could not and would be silly),  the interceptor would have succeeded - the original target is now safe.

@CharlesBretana Against heatseeking weapons at such close range, I doubt that most jets even w/ high off boresight capability would find much benefit using an interceptor missile. However, against radar guided missiles at range, a combination of RWR and high off boresights capability as seen on something like the F35 should give the missile a better chance. Most BVR engagements happen at decent ranges, enough to start notching against older missiles as well as trying to bleed speed away as much as possible

Althoigh I'm not on completely solid ground here, even A/A missile engagements that occur at or near maximum F-pole ranges, (with possible exception of AIM-54, although I am not familiar with current Soviet or other nation technologies), still have less than 30 second TOFs. My experience is from the 70s through 90s, in the F-4, not a current fighter, so I am guessing here, but I would guess that even in a current 6th gen fighter (unless extremely lucky, or using external targeting data), it would take longer than that to acquire, track, etc another A/A missile,

New BVR missiles fly a bit longer, but not more then a couple of minutes. Therefore as stated above, it is possible under good conditions, but unlikely.

@CharlesBretana From my understanding, and you probably know more about this, the BVR capabilities of aircraft and missiles don't necessarily translate into when a semi or active radar guided missile is fired. There are still things like VID which happened during the Gulf Wars. Also w/ the F22, F35 and gen 4++  taking approaches to keep RCS in mind, we're seeing an even heavier emphasis on BVR now that LO is a factor. Going close w/ a LO fighter doesnt make sense. Plus with more advanced sensor fusion engagement distances have opened up.  New AIM 260 and PL15 for example have some hefty reach

All missiles can only intercept a target if it (the target), flies within a specified distance in front of the launch aircraft, referred to as F-pole. F-pole is defined as the slant range distance in front if the launch aircraft that the missile would be after motor burn out and missile deceleration back to launch aircraft TAS, if neither missile nor launch aircraft maneuvers.  Maximum launch range (BVR or visually), is almost always determined primarily by F-pole, but varies enormously with aspect angle, (head-on versus tail chase)

So the missile launch capabilities, based on F-pole range and target aspect angle, along with launch aircraft and target TAS, do play a large role jn determining a missiles launch envelope.  As to the AIM-26O, golly, it's not going to be safe to fly anymore!

When you design a missile, a significant part of the parameters are based on "what are you trying to kill?" The effort needed to disable an aircraft in order to knock it out of the sky is quite different from that needed to destroy a missile.
The processing speed of the computers inside your missile also limit the upper speed that you can track. In the past, designers used to also specify the lower speed limits - the widespread introduction of helicopters put an end to it. During WW2, there were stories of mechanical computers (used for aiming anti-aircraft guns) on board naval vessels being unable to target some torpedo bombers because the aircraft travelled slower than the computers were designed to handle.
This video shows a reconstruction of the shootdown of Malaysian Airlines flight 17 over Ukraine. Missiles do not impact the target as if they were large bullets. Instead they detonate nearby and shred the target with shrapnel. The only aircraft that I am aware of that are designed to survive in an environment full of shrapnel are ground attack aircraft such as the A-10. For these aircraft, fuel consumption is irrelevant. For passenger aircraft, fuel consumption is the most important operating characteristic that airlines measure. As missiles are far more compact than aircraft, they are harder for the shrapnel cloud to affect and will therefore sustain far less damage than human occupied aircraft.


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Why do most military multi function displays use CRT or LCD instead of something like OLED
As I understand most screens for military MFDs utilize LCD screens or used to use CRT screens. OLED however has some advantages over LCDs especially in terms of weight and space. They're generally also thinner.
Why haven't OLEDs seen use in MFDs for military aircraft?
As far as I know only the F35 considered using OLEDs for its helmet mounted display.
I know that OLEDs have a shorter lifespan hour wise, however is that the main reason?

Related

@sophit In some ways yes and no. That answer seems to point to an economical side of things which makes sense for the CRJs. However newer platforms like the F35 and even upgrades for other platforms made a decision to avoid OLEDs for some reason. Especially for things like HMDs.

That's why "related". Anyway I like this and this comments on the topic.

A quick glance at the timeline of the JSF and OLEDs suggests that OLEDs didn't exist when the JSF was developed. The JSF program is from 1993, the very first practical OLED screen appeared in 1999, its diagonal was only 2.7 inch, and what a researcher means by "practical" is not exactly how you or I would understand the term.

Plus, the military is usually risk adverse, in that they don't want to be the first ones to use the latest whiz-bang technology when an older, proven technology will do the job.

Certification cost and speed. Pretty much all technologies are getting adopted at one point, however only after 10 years someone finally certified that technology.

As strange as it sounds, modern military avionics are pretty primitive compared to what’s available in the civilian marketplace, though they once were state of the art. There are several reasons for this. First, most of these systems were developed a solid 10 years ago so they’re not going to take advantage of latest upgrades or changes in technology, unless it would be absolutely necessary. Second if they are going to make those changes it would be in an upgrade package done by the OEM or similar military contractor, not quickly.  The F-35 was developed a solid 17 years ago, and the F-22 traces its beginnings to the early 1990s, when things like OLED displays weren’t available.  On Friday, December 2, 2022, we will get a chance to see the new B-21 Raider when Northrop Grumman rolls it out.  This new stealth bomber, with its open architecture and Agile driven systems software might offer a glimpse of what technologies would be used in a modern fighter developed in the 2020s.  It would be interesting to see if it uses OLED technology in the B-21’s cockpit displays.

Not only were they designed a long time ago, but the military always chooses to make its system reliable above everything else. Even when designing a new system they choose established technologies, not the most modern technologies.

Changing the equipment for something newer means it's a new equipment now. And new equipment has to go through the full certification procedure to be allowed to fly. And that procedure is so expensive that even if you save considerable money on every screen for few decades it still won't cover for the certification costs.
Even if you are designing a brand new system, you go for tried-and-true technology that's well tested and their failure modes are well understood. Implementing promising new tech carries a risk it'll turn out unfit for the purpose. And in that case you're back to the fist paragraph.
On a side note, it's not just aviation. Aviation adds crippling cost of certification like a cherry on the top, but this phenomenon happens everywhere: Governments and military using floppy disks to archive data, NASA using magnetic-core memory as late as the 80's, cars using relays instead of solid-state to control starter motor until recently, Amtrak using rotary converters, etc, etc. When you have good understanding of failure modes, you can simply overengineer that particular aspect or put in procedures to circumvent the problem. Cutting-edge tech is usually reserved for enthusiasts who accept the risk of beta-testing, only then it enters "serious business".
My opinion is that OLEDs being around since barely 1987 are too new to know everything about them and LCDs are not worse enough to make up for the added risk.

new technologies are economic game changers if theyre better, cheaper or more practical.
Military hardware has specific criteria and OLED meet none such standards.
Theyre more expensive, energy intensive and less life expectancy.
And improved Resolution isn't a concern when management of simple data is all the same color.
Green is one best colors to see in any lighting conditions bright or night. Which is why green lasers replaced red.


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How much of the Mig 25 utilized vacuum tubes?
I've often heard that because the Soviet Union lagged behind in integrated circuit development, they built their fighters to rely less on them compared to NATO and American platforms.
The Mig25 is often brought up as an aircraft that utilized a heavy number of vacuum tubes for the aforementioned reason, but also as a defense against EMP related issues when tasked with intercepting American bomber formations.
How much of the Mig25 utilized vacuum tubes, especially in comparison to semiconductor/transistor-based circuitry?

From the Russian sources here and here it is seen that Mig-25 used  vacuum tubes. However these where not the usual large vacuum tubes from consumer electronics, already obsolete at that time. Mig-25 used miniature vacuum tubes specially designed for this aircraft.
Sources say the electronics was anyway efficient and worked well.

Please reproduce the pertinent parts of your links here in the answer itself. Links rot.


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Getting Minimum Grid Area Altitude from navigational chart
One of the questions for the EASA PPL(A) exam sounds like this:

What is the highest Minimum Grid Area Altitude between BERTA (point 1) and LJLJ (Ljubljana) along route A (yellow)?

And the following chart is provided:

The correct answer is 10700 ft, but I'm not sure why and how to detect this from this chart.

The Minimum Grid Area Altitude in each sector is depicted by the truncated numbers shown in the circle below.  (There are several sectors shown in your photo, and the sector further west showing 97 is 9,700' for comparison.  You can see a couple others at the very top of the image as well.)


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Induced drag vs parasite drag understanding
I am having the following graphic from one of the question for the EASA PPL(A) exam.
While I know that:
(1) represents "induced drag", as its decreasing with higher speed
(2) represents "parasite drag", as its increasing with higher speed
But I wonder what the following represent:
(3) Total Drag??? as we know total drag is at its minimum when induced drag and parasite drag are equal?
(4) Form Drag???

3 is definitely the sum of 1 and 2 and might be termed as total drag. Form drag should look like 2. 4 looks like a drag term going very high very fast...

Whatever 4 is, it isn't representative of the same object as 1-3.

As others have said, form drag is a component of parasite drag, it is reasonably a part of 2.  If you want to get subtle, 1 and 2 should cross at the minimum point for 3 -- and of course 3 should be at 2x the 1/2 crossing point.

Looking at the answers to the EASA question bank, nothing is answered by item 4. In other words, it's just thrown in as a red herring.

@tedder42: this sounds like "the" answer

The graph above assumes, (is only valid for), LEVEL flight, i.e. when at 1 G, or when the lift being generated is equal to the aircraft weight. The actual distinction between induced drag and parasite drag is simply a matter of definition. Parasite drag is that drag that the aircraft would experience at zero Lift, at that AOA where the coefficient of Lift is zero, i.e., the drag due only to the aircraft's veloctiy. Induced drag is just, well, everything other than the parasite drag, or the extra drag caused by Lift

The graph above assumes, (is only valid for), LEVEL flight, i.e. when at 1 G, or when the lift being generated is equal to the aircraft weight. The actual distinction between induced drag and parasite drag is simply a matter of definition. Parasite drag is that drag that the aircraft would experience at zero Lift, - at that AOA where the coefficient of Lift is zero, i.e., the drag due only to the aircraft's velocity. Induced drag is just, well, everything other than the parasite drag, or the extra drag caused by Lift.
So, your statment (1) above, to wit,
" ... represents "induced drag", as its decreasing with higher speed" is also only true for 1 G LEVEL flight.
To be accurate it should say:
"Induced drag increases as Angle of Attack increases."
Form drag is a euphonism for parasite drag. Sometimes it is used to separate, (or make a distinction), between the parasite drag from a clean aircraft, and the extra parasite drag created by interference drag between external stores (like wing tanks or bombs), although I've also seen the term Interference Drag used for that purpose.

You are correct about 1, 2 & 3.
As Michael Hall has commented, 4 is not representative of the same object that the other three curves represent. To figure out what 4 represents, let's look at the drag equation:
$$Drag = ½ρV² \cdot C_D \cdot S$$
Now the "speed" in the chart is most probably IAS (dynamic pressure). We observe in the chart that 4 attains a drag equal to 2 at a much lower "speed". This can be a result of two factors: 4 either has a high $C_D$ or a large surface area, or both.
In my opinion, 4 represents a parachute or a similar object. A parachute produces a lot of drag at relatively low speeds, And that's precisely what the object in 4 is doing.

Number 3 is total drag and it is the sum of number 1 and 2, the lowest point of the total drag curve is where minimum drag exists.

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Easy way to calculate distance between two GPS coordinates
I'm studying for my EASA PPL(A) exam, and for some questions for the Navigation exam, I'm requested to calculate the distance between two points, like so:

What is the distance from VOR Brünkendorf (BKD) (53°02?N, 011°33?E) to Pritzwalk (EDBU) (53°11'N, 12°11'E)?
A. 24 km
B. 42 NM
C. 24 NM <--- this is the correct answer
D. 42 km

I'm given also the following attached chart:

What's the best method to apply in order to calculate the distance? Should I apply a formula or "just" by looking at the chart?

One minute of latitude is equal to 1 nautical mile of distance, and one minute of longitude is equal to $\cos({\rm latitude})$ nautical miles. So the distance between two points can be computed to a good approximation as:
$$ \sqrt{(\text{difference in latitude})^2+(\text{difference in longitude}\times \cos({\text{latitude}))^2}} $$
So, in this case:
$$ \sqrt{9^2+(\cos(53^\circ)*38)^2}\approx 24.6~\rm nm $$

Note that this approximation only works well for small distances (which is the case here). For points that are thousands of nautical miles apart, this won't work...

@Bianfable Of course not. But the same thing is true of a chart: the projection used on a chart is pretty much the same approximation, so this method is valid basically any time you would be using a chart to measure distances.

This seems like a lot more work.

One minute of latitude is equal to one minute of latitude, and one nautical mile is equal to 1852 meters. "One minute of latitude" is not a constant measurement because the Earth is not a perfect sphere.

@randomhead Technically true, but that's like saying "the length of one meter stick is the length of one meter stick, and one meter is equal to the distance light travels in a vacuum in 1/299792458 seconds. The length of a meter stick is not a constant measurement because meter sticks expand and contract when the temperature changes and not every meter stick is exactly the same length to begin with." No measuring tool is perfect, but "one minute of latitude equals 1 nautical mile" is plenty accurate for the precision demanded in aviation.

@MichaelHall It's a handful of keypresses on a calculator (14 on mine). Depending on your familiarity with your calculator it may or may not be faster than measuring it with a plotter. In any case it's another option and if you don't have a good straightedge I think it's probably easier than trying to use your fingers. My fingers, at least, do not cooperate and keep the same distance apart as I'm moving them over to the scale.

@Chris, excellent comparison to the length of a meter...  Randomhead apparently thinks it's significant since "they" made the same comment on each of our answers.  Anyway, my technique is easier to remember, and I challenge you to a race any time!

@MichaelHall Ahah, to each their own. I can pound out the numbers on my calculator faster than I can even get the plotter lined up. Probably my physics background ;)

One word:  Measure.
I'm not even sure how you would calculate such a thing, and you can't get the answer by "just" looking at it either.
Use a plotter, or put the straight edge of a piece of paper between the points, mark it with a pencil, and compare it to the scale of miles in the legend of the chart.
Alternately, compare distance to the tick marks on a line of longitudinal.  Each minute of a degree is equal to a nautical mile. (see red lines in picture below for reference.)

On a multiple choice question like this you could probably put your fingers on the screen and estimate close enough to distinguish between 42 and 24 nm.

Yes, I was having that as an option. But the issue is that the exam will be on computer, so the charts will be digitally available. Also, the chart doesn't have any scale on it. Can I assume that the scale of aeronautical charts is usually 1:500000

I added some detail to my answer.

"On a multiple choice question like this you could probably put your fingers on the screen and estimate close enough to distinguish between 42 and 24 nm." Unfortunately, 42 kilometers is 23 nautical miles, which might be too close for finger estimating.

Hi @Chris, I was doing exactly the same, and somehow I guessed it. But was also curious if there is a formula to calculate it. Using a plotter should be fairly simple. Thanks for the suggestion.

@Chris, I didn’t consider that, good point.

Each minute of a degree is equal to one minute, and one nautical mile is equal to 1852 meters. "One minute of latitude" is not a constant measurement because the Earth is not a perfect sphere.

@randomhead, technically that's true.  In practice, the distinction is completely irrelevant.


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Calculating center of gravity when calculated CoG, T/O mass, fuel burn and fuel burn station are known
I'm practicing for my EASA PPL(A) exam and one of the questions I came across sound like this:

Calculated take-off mass = 1082 kg, calculated CG = 0.254 m, fuel burn
= 55 l on station 0.40 m.
A. 24.6
B. 25.2
C. 25.4
D. 24.8   <- this is the correct answer

My question is how do I actually get to this answer?
This is what I tried:

14,6 cm would be the difference between the fuel burn station and calculated CoG (in cm)
1027 kg would be the difference between calculated T/O mass and fuel burned
803 cm/kg would be the removed mass
0,78 cm would be the CoG shift calculated
New CoG should be 25,4 (old CoG) - 0,78 which would mean 24,6

But yeah... that is not correct. So what am I missing here?

As @Bianfable has already explained to you, 55 liters of fuel do not correspond to 55 kilogrammes so your final weight is not 1082-55=1027kg...

but that's the thing, in another question on the same test, when i calculated 55 x 0,72 (density of avgas), the numbers i got were correct... so that's why here I assumed that I should not convert it.

@sophit, so I should calculate 0,72 for avgas density?

@TudorRavoiu: you cannot mix liters with kilogrammes. The former measures a volume, the latter a mass, they are two very different things. 1kg of cast iron is not heavier than 1kg of feathers. The two volumes anyway are going to be very different. Viceversa 1l of air is going to be much lighter than 1l of lead.

Hi Tudor, welcome to aviation.stackexchange. Can you please show your calculations you made, not only the intermediate results?

@TudorRavoiu When the metric system was created, the original definition of one kilogram is "the mass of one liter of water". Now, this is a perfectly fine and workable definition, but it seems to me (granted, this is based on only two data points) that this has created a kind of false equivalence in your head. You're so used to thinking that "kilogram = liter" that you think there's never any need to convert from one to the other. But that equation is ONLY true for pure water (and even then, it's only an approximation, since we've moved on to more precise definitions of units).

You are doing 1082 - 55 = 1027.  The point that people are trying to make is that you have skipped the step you did correctly in the other problems you described.
You want 1082 - (55 * 0.72) = 1,042.4kg.  You are subtracting the 55 as if it is the weight of the fuel in kg, but it is not.  55 liters of AVGAS weighs 39.6kg.  Being a PPL test I will take the liberty of assuming the density required is for AVGAS, and using that density the problem works:

Weight (kg)
Arm (m)
Moment Kg-M

TO Wt/Bal
1082 (given)
0.254 (given)
274.828 (step 1)

Fuel Burn
-39.6 (step 2)
0.4 (given)
-15.84 (step 3)

Ldg Wt/Bal
1042.4 (step 4)
0.248 (step 6)
258.988 (step 5)

Steps:

Moment(TO) = Weight(TO) * Arm(given) = 1082 * 0.254 = 274.828 kg-m
Weight(fuel burn) = Volume(fuel burn) * Density(fuel) = -55l * 0.72 kg/l = -39.6 kg
Moment(fuel burn) = Weight(fuel burn) * arm(fuel) = -39.6 kg * 0.4m = 15.48 kg-m
Weight(ldg) = Weight(TO) - Weight(fuel burn) = 1082 - 39.6 = 1042.4 kg
Moment(ldg) = Moment(TO) - Moment(fuel burn) = 274.828 - 15.48 = 258.988
Resulting CG Arm = Moment(ldg) / Weight(ldg) = 258.988/1042.4 = 0.248m = 24.8cm

24.8 = You pass the test = you become a pilot = jackpot.

Ehm, actually you passed the test

Thank you for taking the time to write this nice answer and actually helping out.

Thank you, I honestly thought about converting to Avgas after an answer from a previous question I had on aviation.stackexchange. But seeing that there is no specification to what kind of plane it's being used, I though... I shouldn't.... Maybe I should not think so much :). Thank you @DeltaLima

"Maybe I should not think so much": don't think, just do (Maverick)